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Question 6: Draw the organic products formed when each alkyne is treated with two equivalents of HBr. a. CH₂CH₂CH₂CH₂-C=C-H b. CH3-C=C-CH₂CH3 C. C=CH

Question 6: Draw the organic products formed when each alkyne is treated with two equivalents of HBr. a. CH₂CH₂CH₂CH₂-C=C-H b. CH3-C=C-CH₂CH3 C. C=CH

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question 6
**Question 5:** Draw the products of the following reaction, including stereochemistry. A chemical reaction is shown: - Reactants: trans-2-butene with Br₂ in the presence of H₂O. - Product box is left empty for students to fill in. **Question 6:** Draw the organic products formed when each alkyne is treated with two equivalents of HBr. **a.** CH₃CH₂CH₂C≡CH **b.** CH₃–C≡C–CH₂CH₂CH₃ **c.** Cyclopentylacetylene (depicted as a 5-membered ring with a triple bond substituent outside the ring) **Question 7:** Draw the products formed when 3-hexyne is treated with each reagent: (a) Br₂ (2 equivalents); (b) Cl₂ (1 equivalent). - **a.** Br₂ + CH₃CH₂C≡CCH₂CH₃ → ????? - **b.** Cl₂ + CH₃CH₂C≡CCH₂CH₃ → ?????
Transcribed Image Text:**Question 5:** Draw the products of the following reaction, including stereochemistry. A chemical reaction is shown: - Reactants: trans-2-butene with Br₂ in the presence of H₂O. - Product box is left empty for students to fill in. **Question 6:** Draw the organic products formed when each alkyne is treated with two equivalents of HBr. **a.** CH₃CH₂CH₂C≡CH **b.** CH₃–C≡C–CH₂CH₂CH₃ **c.** Cyclopentylacetylene (depicted as a 5-membered ring with a triple bond substituent outside the ring) **Question 7:** Draw the products formed when 3-hexyne is treated with each reagent: (a) Br₂ (2 equivalents); (b) Cl₂ (1 equivalent). - **a.** Br₂ + CH₃CH₂C≡CCH₂CH₃ → ????? - **b.** Cl₂ + CH₃CH₂C≡CCH₂CH₃ → ?????
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On 6b, why are the two Br on the 2nd carbon and not the 3rd??

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